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The electronic configuration of Pt (atomic number 78) is:

Structure of Atom · Class 11 · JEE Main Previous Year Question

Question

The electronic configuration of Pt (atomic number 78) is:

Options
  1. a

    [Kr]4f145d10[\text{Kr}]\,4f^{14}\,5d^{10}

  2. b

    [Xe]4f145d10[\text{Xe}]\,4f^{14}\,5d^{10}

  3. c

    [Xe]4f145d86s2[\text{Xe}]\,4f^{14}\,5d^8\,6s^2

  4. d

    [Xe]4f145d96s1[\text{Xe}]\,4f^{14}\,5d^9\,6s^1

Correct Answerd

[Xe]4f145d96s1[\text{Xe}]\,4f^{14}\,5d^9\,6s^1

Detailed Solution

🧠 The Stability Exception Platinum (Z=78Z=78) is infamous in atomic structure. It sits in Period 6, where the expected 5d6s5d-6s energy gap is so small that electrons "relocate" for extra stability. This is the same principle that makes \ceCr\ce{Cr} or \ceCu\ce{Cu} exceptions, but slightly more complex.

🗺️ The Baseline Build

  1. Preceding Noble Gas: Xenon (Z=54Z=54).
  2. Remaining Electrons: 7854=2478 - 54 = 24.
  3. The Shell Fill:
    • 6s26s^2: 2 electrons.
    • 4f144f^{14}: 14 electrons (Complete).
    • 5d85d^8: 8 electrons (Wait!).
  4. The "Shift": In heavy metals like Pt, the 5d5d and 6s6s energy levels are nearly degenerate. For Pt, the most stable state actually pulls an electron from 6s6s to 5d5d, resulting in 5d96s15d^9 6s^1. Result: [Xe]4f145d96s1[\text{Xe}] 4f^{14} 5d^9 6s^1.

The Group 10 Trap Platinum belongs to Group 10 (\ceNi,Pd,Pt\ce{Ni, Pd, Pt}).

  • \ceNi\ce{Ni}: 3d84s23d^8 4s^2 (Standard)
  • \cePd\ce{Pd}: 4d105s04d^{10} 5s^0 (Anomaly)
  • \cePt\ce{Pt}: 5d96s15d^9 6s^1 (Anomaly) Recognizing that the lower members of this specific group are anomalies helps you avoid the "standard" d8s2d^8 s^2 traps.

⚠️ Common Traps Option (2) [Xe]4f145d10[\text{Xe}] 4f^{14} 5d^{10} is a common "over-simplification" guess. Remember, d10d^{10} is completely full. If Pt had that, it wouldn't have any valence electrons to form common +2+2 states. The correctly identified anomaly is the s1s^1 configuration.

Answer: (d)\boxed{\text{Answer: (d)}}

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